Measure, Integration & Real Analysis, 2021a

42 Chapter 2 Measures
• Suppose X is a set, S is a σ-algebra on X, and w : X → [0, ∞] is a function.
Define a measure μ on (X, S) by
μ(E) = ∑ w(x)
x∈E
for E ∈S. [Here the sum is defined as the supremum of all finite subsums
∑ x∈D w(x) as D ranges over all finite subsets of E.]
• Suppose X is a set and S is the σ-algebra on X consisting of all subsets of X
that are either countable or have a countable complement in X. Define a measure
μ on (X, S) by
{
0 if E is countable,
μ(E) =
3 if E is uncountable.
• Suppose S is the σ-algebra on R consisting of all subsets of R. Then the function
that takes a set E ⊂ R to |E| (the outer measure of E) is not a measure because
it is not finitely additive (see 2.18).
• Suppose B is the σ-algebra on R consisting of all Borel subsets of R. We will
see in the next section that outer measure is a measure on (R, B).
The following terminology is frequently useful.
2.56 Definition measure space
A measure space is an ordered triple (X, S, μ), where X is a set, S is a σ-algebra
on X, and μ is a measure on (X, S).
Properties of Measures
The hypothesis that μ(D) < ∞ is needed in part (b) of the next result to avoid
undefined expressions of the form ∞ − ∞.
2.57 measure preserves order; measure of a set difference
Suppose (X, S, μ) is a measure space and D, E ∈Sare such that D ⊂ E. Then
(a) μ(D) ≤ μ(E);
(b) μ(E \ D) =μ(E) − μ(D) provided that μ(D) < ∞.
Proof
Because E = D ∪ (E \ D) and this is a disjoint union, we have
μ(E) =μ(D)+μ(E \ D) ≥ μ(D),
which proves (a). If μ(D) < ∞, then subtracting μ(D) from both sides of the
equation above proves (b).
Measure, Integration & Real Analysis, by Sheldon Axler